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6/11 = 0.5454545454545454....    evey 10th spot is a 4    100th spot is a 4

What real value of t produces the smallest value of the quadratic t^2 -12t - 36?    

I used the slider at Desmos and the smallest value is when t = 6.  

_.

Law of Cosines on triangle PQM

PQ^2 + PM^2 - QM^2 = 2*PQ*PM*cosQPM

Law of Cosines on triangle RPM

PR^ 2+ PM^2 - RM^2 = 2*PR*PM*cosRPM

==> PM = 7*sqrt(3)

. ln(c) - ln(d) = ln(c/d)

so  ln(x-3)^9 - ln(x^11) can be written as  ln( (x-3)^9 / x^11 )    i.e. ln( ( x-3)9 / x11 )

I agree that there are a possible 32! pairings in total, and that the maximum number of possible matches in a draw hosted by the seeded team is 31.  This is the same as saying the minimum number of matches hosted by the unseeded team in a draw is 1.  Since there are 32 unseeded teams the number of possible draws that result in just 1 unseeded team hosting is 32, so I make the probability p = 32/32! or p = 1/31!.

1.  a*ln(b) = ln(b^a)   so 9*ln(x-3) - 11*ln(x) can be written as  ln(x-3)^9 - ln(x^11)

2. ln(c) - ln(d) = ln(c/d)  so  ln(x-3)^9 - ln(x^11) can be written as  ln( (x-3)^9/x^11 )    i.e. ln( ( x-3)⁹x-¹¹ )

(Edited to correct mistaken sign)

First, multiply the whole equation by \((x-3)^2(x+5)\):

\(1 = A(x-3)^2+B(x+5)(x-3)+C(x+5)\\ \)

If we set x to be -5, we make B and C equal to zero, which we can use to solve for A:

\(1=A(-5-3)^2\\ 1=A(-8)^2\\ 1=64A\\ \boxed{A=\frac{1}{64}}\)

To answer your private message question

I put each seeded team number with the unseeded number directly below

32 with 31 etc

the ones left at the end are  seeded 1 with unseeded 32

The maximum number hosted by the seeded team is 31

32 with 31

31 with 30

..

2 with 1           so far all hosted by seeded team

1 with 32         hosted by unseeded team

I cannot see any other way of doing this.

If seeded are lined up and each one randomly paired with an unseeded then there wll be 32*31*.....*1 = 32! possible matches

So I think the prob that 31 games will be hosted by the seeded team is    1/32!

Annie is planning a business meeting for her company. She has a budget of 1,325 for renting a meeting room at a local hotel and providing lunch. She expects 26 people to attend the meeting. The cost of renting the meeting room is 270. Which inequality shows how to find the amount, (x), Annie can spend on lunch for each person?

Hello Guest!

\(26x<1325-270\\ \color{blue}x<(1325-270)/26\)

\(x<40.5\overline{769230}\)

Annie can spend on lunch for each person 40.57.

  !

Let AB and CD be chords of a circle, that meet at point Q inside the circle. If AQ = 6, BQ = 14, and CD = 38, then find the minimum length of CQ.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AQ * QB = CQ * QD

6 * 14 = 84

x*(38 - x) = 84         CQ = 19 - √277 = 2.356683023

It was a bit hard for you to understand when nothing I had written made any sense :)

I finally understand now!

Sorry for being stubborn and not understanding quicker

I appreciate the time you took to help me

Keep it up Melody 

I really don't want to be mean, but this is the standard format of a purpose wrong answer from a guest, anybody that sees this:

"by the some rule, this is the answer"

do not enter the answer

By Descartes Rule of Signs, there are three real soutions.

We try to set the equation to:

$y=mx+b$ as $m$ is the slope

$x=3$

however we notice without a $y$, $m$ is not findable, therefore the slope is undefined.

Why didn't you talk about what you had tried for yourself MathSolver?

There is no evidence that you were trying to solve anything.  So why call yourself a MathSolver?

Looks like you are just asking for other people to solve your homework problem for you.

Refer to the diagram below:

The area of the original rectangle can be assigned any value l since we are only interested in the area ratio of the 2 internal sections.

So I have let the area be 2 units

The width is w

The length is 2/w

The original brown rectangle and the smaller red rectangle are similar figures.  So I have said the ratio of the sides is delta where delta is a positive number less than 1.

From this all the given dimensions and be found.

Are of blue parallelogram \(=\frac{2\delta}{w}*w(1-\delta)=2\delta(1-\delta)\)

Area of green triangle =\( \frac{1}{2}*\frac{2}{w}(1-\delta)*\delta w =\delta(1-\delta) \)

\(\dfrac{\text{Area }TSPA}{\text{Area } PRB} =2\)

[TSPA] / [PRB] = 2

Angle x = 50 degrees

These types of questions are presented in intelligence tests; they never have context beyond the question itself.  This question would be one in a series used to measure pattern recognition skills.  I’ve taken several of these tests, and I’ve not seen this question before; however, the sequence is obvious:

This is a prime number sequence in the form of the first letter of the (English) word for the number.  (T)wo, (T)hree, (F)ive, (S)even, (E)leven, (T)hirteen, (S)eventeen, (N)ineteen, ....

GA

Thank you  for helping me understand!

1. Round them 

10*6*7=420

Your instinct was correct: EG isn't necessarily equal to FG.

Here's a solution:

Notice that DGH and DFE are similar. We can set up an equation:

\(\frac{x-4}{x-7}=\frac{(x-4)+2x}{(x-7)+(x+3)}\\ \frac{x-4}{x-7}=\frac{3x-4}{2x-4}\\ (x-4)(2x-4)=(x-7)(3x-4)\\ 2x^2-12x+16=3x^2-25x+28\\ 0=x^2-13x+12\\ (x-12)(x-1)=0\\ x=1,x=12\)

1 is extraneous because it will be negative for x-4, which is impossible.

Therefore the answer is \(\boxed{24}\)

Using the hint at the very end, we can multiply \(5.5x+7.5y=930\) by -2 to make it easier:

\(-11x-15y=-1860\\ 12x+15y=1920\)

Add the equations together to cancel out the y term:

\(x=60\)

Now that we know x, we can plug x back into any of the 2 equations above to solve for y (I chose the second one because it seems simpler):

\(12(60)+15y=1920\\ 720+15y=1920\\ 15y=1200\\ y=80\)

Therefore, the answer is:

\(\boxed{x=60, y=80}\)